Electromagnetic Induction Test

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Electromagnetic Induction Test - 66

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Question 1
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A flat circular coil of $$n$$ turns, area $$A$$ and resistance $$R$$ is placed in a uniform magnetic field $$B$$. The plane of coil is initially perpendicular to $$B$$. When the coil is rotated through an angle of $${180}^{o}$$ about one of its diameter, a charge $${Q}_{1}$$ flows through the coil. When the same coil after being brought to its initial position, is rotated through an angle of $${360}^{o}$$ about the same axis a charge $${Q}_{2}$$ flows through it. Then $${Q}_{2}/{Q}_{1}$$

A
$$1$$

B
$$2$$

C
$$1/2$$

D
$$0$$

Solution

Net charge flowing through the coil is given by $$Q = \dfrac{\Delta \phi}{R}$$ where $$R$$ is the resistance

Initially the plane of ring is perpendicular to $$B$$, i.e Area vector $$\vec{A}$$ is parallel to $$\vec{B}$$
where $$A$$ is the area of the coil.

$$\therefore $$ initial flux $$ \phi_i = \vec{A}.\vec{B} = AB$$

Case 1) : Coil is rotated by $$180^o$$ i.e $$\vec{A}$$ $$\parallel$$ $$-\vec{B}$$

$$\therefore$$ Final flux $$\phi_f = -AB$$

$$\implies |\Delta \phi | = |\phi_f - \phi_i| = 2AB$$

Thus $$Q_1 = \dfrac{2AB}{R}$$ .............(1)

Case 2): Coil is rotated by $$360^o$$, i.e $$\vec{A} $$ $$ \parallel$$ $$\vec{B}$$

Thus final flux $$\phi_f = AB$$

$$\therefore \Delta \phi = \phi_f - \phi_i = 0$$

Hence $$Q_2 = \dfrac{\Delta \phi}{R} = 0$$

$$\implies \dfrac{Q_2}{Q_1} = 0 $$
Question 2
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A rectangular loop of sides $$a$$ and $$b$$ is placed in $$x-y$$ plane. A uniform but time varying magnetic field of strength $$\vec { B } =20t\hat { i } +10{ t }^{ 2 }\hat { j } +50\hat { k } $$ is present in the region. The magnitude of induced emf in the loop at time $$t$$ is :

A
$$20+20t$$

B
$$20$$

C
$$20t$$

D
zero

Solution

$$\vec { S } =(ab)\hat { k } $$ $$\rightarrow$$ perpendicular to x-y plane
$$\phi =\vec { B } \cdot \vec { S } =(50)(ab)=$$ constant
$$\cfrac { d\phi }{ dt } =0$$
$$\therefore \quad e=0$$
Question 3
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A square loop of side $$b$$ is rotated in a constant magnetic field $$B$$ at angular frequency $$\omega$$ as shown in figure. What is the emf induced in it?

A
$${b}^{2}B\omega \sin { \omega t } $$

B
$$bB\omega \sin ^{ 2 }{ \omega t } $$

C
$$b{B}^{2}\omega \cos { \omega t } $$

D
$${b}^{2}B\omega$$

Solution

At time $$t$$, angle rotated by loop is $$\theta=\omega t$$. This is also the angle between $$\vec { B } $$ and $$\vec { S } $$. Then
$$\phi=BS\cos { \theta } =B{b}^{2}\cos { \omega t } $$
$$e=\left| \cfrac { d\phi }{ dt } \right| ={b}^{2}B\omega \sin { \omega t } $$
Question 4
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In the figure shown, a uniform magnetic field $$\left| \vec { B } \right| =0.5T$$ is perpendicular to the plane of circuit. The sliding rod of length $$l=0.25m$$ moves uniformly with constant speed $$v=4 m{s}^{-1}$$. If the resistance of the slides is $$2\Omega$$, then the current flowing through the sliding rod is :

A
$$0.1A$$

B
$$0.17A$$

C
$$0.08A$$

D
$$0.03A$$

Solution

$$e=Bvl$$
$$=0.5\times 4\times 0.25=0.5V$$
$$12\Omega$$ and $$4\Omega$$ are parallel. hence their net resistance $$R=3\Omega$$
$$i=\cfrac{e}{R+r}=\cfrac{0.5}{3+2}=0.1A$$
Question 5
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In the figure magnetic field points into the plane of paper and the conducting rod of length $$l$$ is moving in this field such that the lowest point has a velocity $${v}_{1}$$ and the topmost point has the velocity $${v}_{2}({v}_{2}> {v}_{1})$$. The emf induced is given by :

A
$$B{v}_{1}l$$

B
$$B{v}_{2}l$$

C
$$\cfrac { 1 }{ 2 } B({v}_{2}+{v}_{1})l$$

D
$$\cfrac { 1 }{ 2 } B({v}_{2}-{v}_{1})l$$

Solution

Velocity of a point at a distance $$x$$ from the bottom is given by:
$$v={v}_{1}+\left( \cfrac { { v }_{ 2 }-{ v }_{ 1 } }{ l } \right) x$$
Potential difference on a small length at this distance is $$e=Bv(dx)$$
$$\therefore$$ Total potential difference $$e=\int _{ 0 }^{ l }{ Bvdx } $$
$$\displaystyle e=\int_0^l {B \left(v_1+\dfrac{(v_2-v_1)x}{l} \right)dx}$$
$$=\dfrac{B(v_1+v_2)l}{2}$$
Question 6
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Two parallel long straight conductors lie on a smooth plane surface. Two other parallel conductors rest on them at right angles so as to form a square of side $$a$$. A uniform magnetic field $$B$$ exists at right angles to the plane containing the conductors. Now conductors start moving outward with a constant velocity $${v}_{0}$$ at $$t=0$$. Then induced current in the loop at any time $$t$$ is ($$\lambda$$ is resistance per unit length of the conductors) :

A
$$\cfrac { aB{ v }_{ 0 } }{ \lambda (a+{ v }_{ 0 }t) } $$

B
$$\cfrac { aB{ v }_{ 0 } }{ 2\lambda } $$

C
$$\cfrac { B{ v }_{ 0 } }{ \lambda } $$

D
$$\cfrac { B{ v }_{ 0 } }{ 2\lambda } $$

Solution

At $$t=t$$ side of square,
$$l=(a+2{v}_{0}t)$$
Area $$S={l}^{2}={(a+2{v}_{0}t)}^{2}$$
$$\phi=BS=B{(a+2{v}_{0}t)}^{2}$$
$$e=\cfrac{d\phi}{dt}=4B{v}_{0}(a+2{v}_{0}t)$$
$$R=\lambda[4l]=4\lambda(a+2{v}_{0}t)$$
$$\therefore$$ $$i=\cfrac{e}{R}=\cfrac{B{v}_{0}}{\lambda}$$
Question 7
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Two metallic rings of radius $$R$$ are rolling on a metallic rod. A magnetic field of magnitude $$B$$ is applied in the region. The magnitude of potential difference between points $$A$$ and $$C$$ on the two rings (as shown), will be :

A
$$0$$

B
$$4B\omega{R}^{2}$$

C
$$8B\omega{R}^{2}$$

D
$$2B\omega{R}^{2}$$

Solution

$${V}_{A}-{V}_{0}=\cfrac { B\omega { \left( 2R \right) }^{ 2 } }{ 2 } =2B\omega { R }^{ 2 }......(i)$$
$${V}_{0}-{V}_{C}=\cfrac { B\omega { \left( 2R \right) }^{ 2 } }{ 2 } =2B\omega { R }^{ 2 }......(ii)$$
Adding these tow equations we get,
$${V}_{A}-{V}_{C}=4B\omega {R}^{2}$$
Question 8
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A square loop of side $$a$$ and a straight long wire are placed in the same plane as shown in figure. The loop has a resistance $$R$$ and inductance $$L$$. The frame is turned through $${18}^{o}$$ about the axis $$OO'$$. What is the electric charge that flows through the loop?

A
$$\cfrac { { \mu }_{ 0 }Ia }{ 2\pi R } \ln { \left( \cfrac { 2a+b }{ b } \right) } $$

B
$$\cfrac { { \mu }_{ 0 }Ia }{ 2\pi R } \ln { \left( \cfrac { b }{ { b }^{ 2 }-{ a }^{ 2 } } \right) } $$

C
$$\cfrac { { \mu }_{ 0 }Ia }{ 2\pi R } \ln { \left( \cfrac { a+2b }{ b } \right) } $$

D
None of these

Solution

$${ \phi }_{ i }=\int _{ b }^{ b+a }{ \cfrac { { \mu }_{ 0 } }{ 2\pi } \cfrac { i }{ x } } (adx)$$

$$\quad =\cfrac { { \mu }_{ 0 }ia }{ 2\pi } \ln { \left( \cfrac { b+a }{ a } \right) } $$
Similarly

$${ \phi }_{ f }=\cfrac { { \mu }_{ 0 }ia }{ 2\pi } \ln { \left( \cfrac { b-a }{ a } \right) } $$

$$\Delta \phi =\left| { Q }_{ i }-{ Q }_{ f } \right| =\cfrac { { \mu }_{ 0 }ia }{ 2\pi } \ln { \left( \cfrac { b+a }{ b-a } \right) } $$

$$\quad \Delta q=\cfrac { \Delta \phi }{ R } =\cfrac { { \mu }_{ 0 }ia }{ 2\pi R } \ln { \left( \cfrac { b+a }{ b-a } \right) } $$
Question 9
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A vertical ring of radius $$r$$ and resistance $$R$$ falls vertically. It is in contact with two vertical rails which are joined at the top. The rails are without friction and resistance. There is a horizontal uniform magnetic field of magnitude $$B$$ perpendicular to the plane of the ring and the rails. When the speed of the ring is $$v$$, the current in the top horizontal of the rail section is

A
$$0$$

B
$$\cfrac{2Brv}{R}$$

C
$$\cfrac{4Brv}{R}$$

D
$$\cfrac{8Brv}{R}$$

Solution

$$Upper\quad part\quad of\quad the\quad rring\quad has\quad resistance\quad R/2\quad and\quad lower\quad part\quad has\quad resistance\quad R/2\\ and\quad they\quad are\quad connected\quad in\quad parallel,\quad thus,\quad effective\quad resistance\quad =R/4\\ i=\dfrac { EMF }{ Resistance } =\dfrac { Bv(2r) }{ R/4 } =\dfrac { 8Bvr }{ R } $$
Question 10
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A flip coil consists of $$N$$ turns of circular coils which lie in a uniform magnetic field. Plane of the coils is perpendicular to the magnetic field as shown in figure. The coil is connected to a current integrator which measures the total charge passing through it. The coil is turned through $${180}^{o}$$ about the diameter. The charge passing through the coil is

A
$$\cfrac{NBA}{R}$$

B
$$\cfrac{\sqrt {3}NBA}{2R}$$

C
$$\cfrac{NBA}{\sqrt {2}R}$$

D
$$\cfrac{2NBA}{R}$$

Solution

The emf induced in the coil is given by $$\dfrac{d\phi}{dt}$$.
The current passing through the circuit thus becomes $$i=\dfrac{\dfrac{d\phi}{dt}}{R}$$.
Hence the total charge that passed through the current integrator=$$\int i dt$$

$$=\dfrac{1}{R}\int \dfrac{d\phi}{dt}dt$$

$$=\dfrac{\Delta \phi}{R}$$

$$=\dfrac{NBA-(-NBA)}{R}$$

$$=\dfrac{2NBA}{R}$$